| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining basic trigonometry (cosine rule) with standard sector formulas. Part (i) is given as 'show that', making it easier. Parts (ii) and (iii) require routine application of sector area and arc length formulas with simple arithmetic. No novel insight required, just methodical application of standard C2 techniques. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(AC = 7.90\) cm | M1, A1, A1 | Attempt to use the cosine formula. Correct unsimplified expression. Show the given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Hence shaded area \(= 22.1\) cm² | M1, M1, A1 | Attempt area of sector using \(\frac{1}{2}r^2\theta\). Attempt area of \(\triangle ACD\), using \((\frac{1}{2})^2 \sin\theta\), or equiv. Obtain 22.1 |
| (iii) (arc) \(DC = 7.90 \times 1.7 = 13.4\) | M1, A1 | Use \(r\theta\) to attempt arc length. Obtain 13.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Hence perimeter \(= 25.3\) cm | M1, A1 | Attempt length of line \(DC\) using cosine rule or equiv. Obtain 25.3 |
**(i)** $AC^2 = 11^2 + 8^2 - 2 \times 11 \times 8 \times \cos 0.8$
$= 62.3796\ldots$
Hence $AC = 7.90$ cm | M1, A1, A1 | Attempt to use the cosine formula. Correct unsimplified expression. Show the given answer correctly | 3 marks
**(ii)** Area of sector $= \frac{1}{2} \times 7.90^2 \times 1.7 = 53.0$
Area of triangle $= \frac{1}{2} \times 7.90^2 \times \sin 1.7 = 30.9$
Hence shaded area $= 22.1$ cm² | M1, M1, A1 | Attempt area of sector using $\frac{1}{2}r^2\theta$. Attempt area of $\triangle ACD$, using $(\frac{1}{2})^2 \sin\theta$, or equiv. Obtain 22.1 | 3 marks
**(iii)** (arc) $DC = 7.90 \times 1.7 = 13.4$ | M1, A1 | Use $r\theta$ to attempt arc length. Obtain 13.4 |
(line) $DC^2 = 7.90^2 + 7.90^2 - 2 \times 7.90 \times 7.90 \times \cos 1.7$
$DC = 11.9$
Hence perimeter $= 25.3$ cm | M1, A1 | Attempt length of line $DC$ using cosine rule or equiv. Obtain 25.3 | 4 marks
**Total: 10 marks**
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{367db494-294e-4b53-b9e8-fd2a69fb6069-3_476_1018_1000_566}
The diagram shows a triangle $A B C$, and a sector $A C D$ of a circle with centre $A$. It is given that $A B = 11 \mathrm {~cm} , B C = 8 \mathrm {~cm}$, angle $A B C = 0.8$ radians and angle $D A C = 1.7$ radians. The shaded segment is bounded by the line $D C$ and the arc $D C$.\\
(i) Show that the length of $A C$ is 7.90 cm , correct to 3 significant figures.\\
(ii) Find the area of the shaded segment.\\
(iii) Find the perimeter of the shaded segment.
\hfill \mbox{\textit{OCR C2 2006 Q7 [10]}}